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Domains and ranges of the inverse hyperbolic functions
Function Domain Range
sinh −1 x ( , ) ( , )
cosh −1 x ( 1 , ) [ 0 , )
tanh −1 x ( −1 , 1 ) ( , )
coth −1 x ( , −1 ) ( 1 , ) ( , 0 ) ( 0 , )
sech −1 x ( 0 , 1 ) [ 0 , )
csch −1 x ( , 0 ) ( 0 , ) ( , 0 ) ( 0 , )

The graphs of the inverse hyperbolic functions are shown in the following figure.

This figure has six graphs. The first graph labeled “a” is of the function y=sinh^-1(x). It is an increasing function from the 3rd quadrant, through the origin to the first quadrant. The second graph is labeled “b” and is of the function y=cosh^-1(x). It is in the first quadrant, beginning on the x-axis at 2 and increasing. The third graph labeled “c” is of the function y=tanh^-1(x). It is an increasing function from the third quadrant, through the origin, to the first quadrant. The fourth graph is labeled “d” and is of the function y=coth^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. The fifth graph is labeled “e” and is of the function y=sech^-1(x). It is a curve decreasing in the first quadrant and stopping on the x-axis at x=1. The sixth graph is labeled “f” and is of the function y=csch^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis.
Graphs of the inverse hyperbolic functions.

To find the derivatives of the inverse functions, we use implicit differentiation. We have

y = sinh −1 x sinh y = x d d x sinh y = d d x x cosh y d y d x = 1 .

Recall that cosh 2 y sinh 2 y = 1 , so cosh y = 1 + sinh 2 y . Then,

d y d x = 1 cosh y = 1 1 + sinh 2 y = 1 1 + x 2 .

We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiation formulas are summarized in the following table.

Derivatives of the inverse hyperbolic functions
f ( x ) d d x f ( x )
sinh −1 x 1 1 + x 2
cosh −1 x 1 x 2 1
tanh −1 x 1 1 x 2
coth −1 x 1 1 x 2
sech −1 x −1 x 1 x 2
csch −1 x −1 | x | 1 + x 2

Note that the derivatives of tanh −1 x and coth −1 x are the same. Thus, when we integrate 1 / ( 1 x 2 ) , we need to select the proper antiderivative based on the domain of the functions and the values of x . Integration formulas involving the inverse hyperbolic functions are summarized as follows.

1 1 + u 2 d u = sinh −1 u + C 1 u 1 u 2 d u = sech −1 | u | + C 1 u 2 1 d u = cosh −1 u + C 1 u 1 + u 2 d u = csch −1 | u | + C 1 1 u 2 d u = { tanh −1 u + C if | u | < 1 coth −1 u + C if | u | > 1

Differentiating inverse hyperbolic functions

Evaluate the following derivatives:

  1. d d x ( sinh −1 ( x 3 ) )
  2. d d x ( tanh −1 x ) 2

Using the formulas in [link] and the chain rule, we obtain the following results:

  1. d d x ( sinh −1 ( x 3 ) ) = 1 3 1 + x 2 9 = 1 9 + x 2
  2. d d x ( tanh −1 x ) 2 = 2 ( tanh −1 x ) 1 x 2
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Evaluate the following derivatives:

  1. d d x ( cosh −1 ( 3 x ) )
  2. d d x ( coth −1 x ) 3
  1. d d x ( cosh −1 ( 3 x ) ) = 3 9 x 2 1
  2. d d x ( coth −1 x ) 3 = 3 ( coth −1 x ) 2 1 x 2
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Integrals involving inverse hyperbolic functions

Evaluate the following integrals:

  1. 1 4 x 2 1 d x
  2. 1 2 x 1 9 x 2 d x

We can use u -substitution in both cases.

  1. Let u = 2 x . Then, d u = 2 d x and we have
    1 4 x 2 1 d x = 1 2 u 2 1 d u = 1 2 cosh −1 u + C = 1 2 cosh −1 ( 2 x ) + C .
  2. Let u = 3 x . Then, d u = 3 d x and we obtain
    1 2 x 1 9 x 2 d x = 1 2 1 u 1 u 2 d u = 1 2 sech −1 | u | + C = 1 2 sech −1 | 3 x | + C .
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Evaluate the following integrals:

  1. 1 x 2 4 d x , x > 2
  2. 1 1 e 2 x d x
  1. 1 x 2 4 d x = cosh −1 ( x 2 ) + C
  2. 1 1 e 2 x d x = sech −1 ( e x ) + C
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Applications

One physical application of hyperbolic functions involves hanging cables . If a cable of uniform density is suspended between two supports without any load other than its own weight, the cable forms a curve called a catenary    . High-voltage power lines, chains hanging between two posts, and strands of a spider’s web all form catenaries. The following figure shows chains hanging from a row of posts.

An image of chains hanging between posts that all take the shape of a catenary.
Chains between these posts take the shape of a catenary. (credit: modification of work by OKFoundryCompany, Flickr)

Hyperbolic functions can be used to model catenaries. Specifically, functions of the form y = a cosh ( x / a ) are catenaries. [link] shows the graph of y = 2 cosh ( x / 2 ) .

This figure is a graph. It is of the function f(x)=2cosh(x/2). The curve decreases in the second quadrant to the y-axis. It intersects the y-axis at y=2. Then the curve becomes increasing.
A hyperbolic cosine function forms the shape of a catenary.

Using a catenary to find the length of a cable

Assume a hanging cable has the shape 10 cosh ( x / 10 ) for −15 x 15 , where x is measured in feet. Determine the length of the cable (in feet).

Recall from Section 6.4 that the formula for arc length is

Arc Length = a b 1 + [ f ( x ) ] 2 d x .

We have f ( x ) = 10 cosh ( x / 10 ) , so f ( x ) = sinh ( x / 10 ) . Then

Arc Length = a b 1 + [ f ( x ) ] 2 d x = −15 15 1 + sinh 2 ( x 10 ) d x .

Now recall that 1 + sinh 2 x = cosh 2 x , so we have

Arc Length = −15 15 1 + sinh 2 ( x 10 ) d x = −15 15 cosh ( x 10 ) d x = 10 sinh ( x 10 ) | −15 15 = 10 [ sinh ( 3 2 ) sinh ( 3 2 ) ] = 20 sinh ( 3 2 ) 42.586 ft .
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Practice Key Terms 1

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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